Leetcode: Generate Parentheses

题目：
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

“((()))”, “(()())”, “(())()”, “()(())”, “()()()”

思路分析：
最后的组合结果总有这样的规律：左括号的个数大于等于右括号的个数。所以，按照这个规律：假设在位置k剩余left个左括号和right个右括号，如果left>0，可以直接打印左括号。，如果left<right且right!=0，则可以打印右括号。如果left和right均为零，则说明已经完成一个合法排列，可以将其打印出来。

C++参考代码：

class Solution
{
private:
    void generateSub(vector<string> &result, string current, int left, int right)
    {
        if (left == 0 && right == 0) result.push_back(current);
        else
        {
            if (left != 0) generateSub(result, current + '(', left - 1, right);
            if (left < right && right != 0) generateSub(result, current + ')', left, right - 1);
        }
    }

public:
    vector<string> generateParenthesis(int n)
    {
        vector<string> result;
        if (n > 0) generateSub(result, string(), n, n);
        return

C#参考代码：

public class Solution
{
    private void GenerateSubString(IList<string> result, string current, int left, int right)
    {
        String value = current;
        if (left == 0 && right == 0) result.Add(value);
        else
        {

            if (left != 0) GenerateSubString(result, current + '(', left - 1, right);
            if (left < right && right != 0) GenerateSubString(result, current + ')', left, right - 1);
        }
    }

    public IList<string> GenerateParenthesis(int n)
    {
        IList<string> result = new List<string>();
        if (n > 0) GenerateSubString(result, "", n, n);
        return